I need help with Stats

Question

A personality scale has scores that are normally distributed with a mean of 150 and a standard deviation of 20. 1. What proportion of people score lower than a percentile rank=143 P(X<143)? 2. What proportion of people score between 157 and 181? P(157<X<181)? 3. What score represents the 60th percentile?

Answer 1

1. What proportion of people score lower than a percentile rank=143 P(X<143)?

 

2. What proportion of people score between 157 and 181? P(157<X<181)?

 

3. What score represents the 60th percentile?

 

ANSWERS:

1.  36.32%  using NORMALCDF(-99999,143,150,20) command in DISTR menu of Ti-83/84 calculator

2.  30.26%  using NORMALCDF(157,181,150,20) command

3.  x = 155  because INVNORM(0.60) = z score .25335; equating this to (x-150)20 ⇒ x = 155.0669

Answer 2

Hi Mya,

 

 

To answer any of these questions we need to calculate z, where

z = (x-μ)/σ

μ is the mean

σ is the standard deviation

 

 

1) P(x<143)

So let's calculate z when x=143

z = (143-150)/20 = -7/20 = -0.35

Look up z=-0.35 in a standard normal table and we have 0.3632 or 36.32%

 

2) P(157<x<181)

We need to calculate the z values at both endpoints.

at 157

z = (157-150)/20 = 7/20 = 0.35

at 0.35, the table gives us 0.6368

at 181

z = (181-150)/20 = 31/20 = 1.55

at 1.55, the table gives us 0.9394

 

Each of these values is the probability that x is less than the given x.  To find the probability that x is between the two values, we subtract

0.9394 - 0.6368 = 0.3026 or 30.26%

 

3) The 60th percentile means look for 60% or 0.600 in the body of the z-table.  Reading the column and row headings for 0.60, we get z = 0.253

z = 0.253 = (x-150)/20

and we solve for x

The score is 155